[paulj000]

Hi,

What is the easiest way to get the current file name in the users URL bar as a variable without the domain and sub directories?

Like if they are at http://www.mydomain.com/dir1/dir2/2003_a.php

then

$currentFile = "2003_a.php"

thanks

[ChristGuy]

Greetingz...

PHP Code:

  $currentFile = $_SERVER["SCRIPT_NAME"]; 


Hope that helps...

[paulj000]

Hi ChristGuy

This is what I tried out:

PHP Code:

<?
$currentFile = $_SERVER["SCRIPT_NAME"];
print $currentFile;
?>

On one of the servers I use this was the output:
/cgi-bin/php

On the other server it output:
/test/03.php
where "test" was one directory deep from site root.

I can delete the directory info if I need to on the second one but I have no idea how to handle the first output!

Any ideas?

[ChristGuy]

You could try using explode...

PHP Code:

  $currentFile = $_SERVER["SCRIPT_NAME"];
  $parts = Explode('/', $currentFile);

  print $parts[count($parts)]; 


That should work...

[paulj000]

Sorry, Let me try to be a little more clear. I am looking for a script that will output only "03.php".

The first url was "http://www.domain1.com/test/03.php"
Output was simply "/cgi-bin/php"

The second url was "http://ww.domain2.com/test/03.php"
Output was "/test/03.php"

How should the script be to output only "03.php" -- on both servers?

Thanks ChristGuy

[ChristGuy]

Greetinz...

If you use the second script what are the results??

And if you use it in a different file on server 1 what do you get?

[kevin]

Hi !

Hope this help

$url ="http://yahoo.com/abcdef/2153abc.gif";

preg_match("/[^\/]+$/",$url,$matches);

$file_name = $matches[0];

print "$file_name";

it will print 2153abc.gif

Regard
Kevin

[paulj000]

Hi ChirstGuy,

The output on both servers was nothing. The page was blank when I used the following code only:

PHP Code:

<?
$currentFile = $_SERVER["SCRIPT_NAME"];
  $parts = Explode('/', $currentFile);

  print $parts[count($parts)];
?>

----------------

Hi Kevin,

I need to be able to have the script get the current file name without supplying the url. The script has to determine what the URL is on it's own and then break it down and output the current page/file name.

Thanks guys

[amailer]

PHP Code:

<?

$currentFile = $_SERVER["SCRIPT_NAME"];

$img = array_pop(explode("/", $currentFile));

echo $img;

?>

see if that works :/

[Man Down]

Christ Guys code will work if you change it to:

PHP Code:

<?
$currentFile = $_SERVER["SCRIPT_NAME"];
  $parts = Explode('/', $currentFile);

  print $parts[count($parts) - 1];
?>